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- Thread starter pan90
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- Aug 30, 2012

- 1,120

You need to show what you have tried, even if it is wrong. That will enable us to help you better.Answer is given, but no explanation or logic for it.

From HiSet free practice test

Note that, for a polynomial with real coefficients, if we have a complex root then we also have the complex conjugate as a root as well. So your cubic polynomial is \(\displaystyle f(x) = a( x - (-1))(x - (2i))(x - (-2i))\).

(The "a" is there because the polynomial will have the same roots no matter what constant is multiplying the whole thing. Your answer key is setting a = 1.)

-Dan

Edit: The method worked to give the right answer, but the signs were wrong. The terms for the given roots are of the form x - r, not x + r as I had originally written.

- Apr 22, 2018

- 251

Right, let’s do it one at a time. Start with $x=-1$. Substitute this in each of the five expressions A–E. You’ll find that B and D give you $0$ whereas A, C and E don’t. Therefore the correct answer is either B or D (you have eliminated A, C and E).

Now substitute $x=2i$ in B and in D. One will give you $0$ while the other won’t. The one that gives you $0$ is the one you’re looking for.

- Jan 29, 2012

- 1,151